如图△ABC中,AB>AC,边AB上去一点D,在变AC上取一点E,使AD等于AE,直线DE的延长线与BC延长线交于点P,求证:BP:CP=BD:CE

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如图△ABC中,AB>AC,边AB上去一点D,在变AC上取一点E,使AD等于AE,直线DE的延长线与BC延长线交于点P,求证:BP:CP=BD:CE如图△ABC中,AB>AC,边AB上去一点D,在变A

如图△ABC中,AB>AC,边AB上去一点D,在变AC上取一点E,使AD等于AE,直线DE的延长线与BC延长线交于点P,求证:BP:CP=BD:CE
如图△ABC中,AB>AC,边AB上去一点D,在变AC上取一点E,使AD等于AE,直线DE的延长线与BC延长线交于点P,求证:BP:CP=BD:CE

如图△ABC中,AB>AC,边AB上去一点D,在变AC上取一点E,使AD等于AE,直线DE的延长线与BC延长线交于点P,求证:BP:CP=BD:CE
证明:
作CF‖AB,交PD于点F
则∠CFE=∠ADE,∠AED=∠CEF
∵AD=AE
∴∠ADE=∠AED
∴∠CEF=∠CFE
∴CE=CF
∵CF‖BD
∴BP∶CP=BD∶CF
∵CE=CF
∴BP∶CP=BD∶CE